- Hey, everyone.
My name is Sahil Chawla.
I'm an MCAT tutor here at Shemmassian Academic Consulting
Today, we'll be going over an organic chemistry passage,
and I'll discuss some of the strategies you can use
when encountering questions like these on the MCAT.
(whooshing)
(electronic beats)
(typing)
All right, so I've got the passage
pulled up here in front of me,
basically just gonna read it out loud,
think out loud so you know exactly what's going on
in my head when I'm reading through such a passage,
and then we'll also go over the questions, as well.
For test day and just for your own independent practice,
you should be going at a pace
around eight minutes per passage.
Since I'm gonna be walking through this,
we're gonna be going a little bit slower than that.
But, for your own independent practice,
shoot for around eight minutes per passage.
All right, so with that all out of the way,
let's go ahead and start with our first paragraph.
In a new study,
researchers aim to synthesize diindolylmethane,
definitely just butchered that,
but we'll go with what they give us here in parentheses,
or DIM, derivatives base thiadiazole
as a new class of urease inhibitors.
DIM is a natural product alkaloid
used extensively in medicinal chemistry.
Alrighty, so just to really help myself engage
with the text, I like to use the outlining strategy
to kinda take what I'm reading and put it into my own words.
So, they talk about these, you know, DIM derivatives here
and that they are urease inhibitors.
Urease is an enzyme
that basically is involved with the hydrolysis of urea.
So, it looks like DIM is gonna be inhibiting
that reaction by inhibiting urease.
So, just to summarize this, I think, really simple,
just DIM, urease, I think that'll suffice.
All right, let's go ahead
and dive into the second paragraph.
Researchers first synthesized
several DIM-based thiadiazole analogs, one through 18,
and characterized them by various spectroscopic techniques,
such as 1H NMR, 13C NMR, and EIMS.
Then, the analogs were evaluated for urease,
parentheses, they give us jack bean urease,
inhibitory potential.
All compounds showed
excellent to moderate inhibitory potential
with an IC50 value within the range of .5 plus or minus .01
to 33.2 plus or minus 1.2 micromolars
compared to what the standard thiourea,
parentheses, they give us 21.6 plus or minus .7 micromolars.
And, then, the synthetic pathway for the analogs
is shown in Figure 1 right here.
Alrighty, so after reading that paragraph, we can also see,
yep, this is the pathway where we're forming these analogs,
right, compounds one through 18.
So, a lot is really discussed there,
but, just to really break it down,
what I'm getting is that there are these DIM analogs, right,
that they had synthesized.
Then, they analyzed them using spectroscopic techniques,
right, H NMR, NMR, right, so I would put spectroscopy.
And, lastly, they just kinda talked about
the inhibitory potential, right?
So, I'll just put inhibitory potential,
just kinda using my own little shorthand here.
I know what that means
and, since this outline is kinda for me,
that's gonna be okay.
So, you shorthand in your outlines, as well,
just to save some time.
Just make sure you remember
what the words meant that you wrote down.
Alrighty, and so I like that
as my outline for paragraph two.
So, let's go ahead and go to paragraph three.
Compound 8, or IC50, which is equal to .5,
or I should say, sorry, compound 8, and, in parentheses,
they give us IC50 equal to .5 plus or minus .01 micromolars,
close parentheses, was the most potent inhibitor
amongst all derivatives.
In the next step, researchers used x-ray crystallography
to determine structure activity relationships
for the compounds.
Researchers discovered that compound 8
forms a pi-stacking interaction
with a conserved Y87 residue within urease.
Mutation of Y87 to alanine
decreases compound 8 potency by 3.7-fold.
The key binding interactions of most active compounds
with enzyme were confirmed
through molecular docking studies.
All right, another paragraph with a lot of information.
What I see they're talking about
is that we have this compound of interest,
which is compound 8.
And, I really, you know, seeing this right here,
this Y87 residue, was really, really stood out to me
because amino acids are such a high-yield topic.
So, when I see 'em, you know, acids appearing in a passage,
that's definitely something I wanna take note of.
You'll see amino acids
not only in the bio and biochem section,
but also in the chem phys section.
That's just how high-yield amino acids are
and even biochemistry in general.
And, on my test day, I even saw biochemistry-type questions
in my CP section of my exam.
So, for me, what's really important here
is that we had this compound 8 and that,
when there was this mutation into alanine
for the 87 position, right,
that there was this 3.7-fold decrease.
So, if I had to summarize this,
I would write C8, Y87A, decreased 3.7-fold.
And, in case you're not aware,
the notation for writing out like wild type or mutations
for amino acids is basically just you put down
the wild-type amino acid, it's position number,
and then the mutant.
Right, so if we look at this one right here.
Y, that's our wild type, 87 is our position number,
and then alanine here, our A, that's the mutant.
Just bringing this up
because you'll definitely see this right here a good amount
with amino acids, again, being such a high-yield topic.
Mutations with amino acids,
you'll see them written out like this.
So, it's good to know, you know,
which one is the wild type, which one is the mutant.
All right, so now with the passage out of the way,
let's go ahead and dive into the questions.
So, question one,
which of the following is not a strong acid.
So, when I see not, least, except,
words like that in a question stem,
I always like to highlight them
just because the MCAT's such an exhausting exam.
It's very easy to like overlook a small detail like that,
and then you might spend a lot of time
digging through the answer choices
and looking for which one is a strong acid, right?
So, I like to just, you know, highlight it on my exam
for not, except, least, words like that.
So, the answer choices they give us
are A, hydrochloric acid, so that's HCl.
They give us acetic acid, so CH3, COOH,
hydrobromic acid, I see HBr,
and then sulfuric acid, which is H2SO4.
So, essentially, for this question
you really just need to know what are your strong acids?
So, I know HCl, HBr, and H2SO4 are our strong acids.
So, acetic acid is gonna be the right answer here.
Acetic acid is known as a weak acid.
So, B will be our right answer.
Make sure you are familiar with the strong acids.
The other ones are gonna be like perchloric acid, so HClO4,
you're gonna have hydroiodic acids, so HI,
and then you'll have HNO3, which is nitric acid.
So, make sure you know these strong acids.
Acetic acid is not one of them.
So, again, B is our best answer.
Let's go ahead and dive into question two.
During the first step of the reaction in Figure 1,
alright, so the first step,
which of the following functional groups present
on the reactants is, again, not present in the product.
So, in the first step of the reaction, so this right here,
which of these functional groups
is found here in the reactants but not here in the products.
So, aromatic rings.
Well, I see them right here, and I also see them right here.
So, these right here, so these are our aromatic rings.
So, I see a lot of those,
and, even as the reaction continues, they basically just,
they're there the whole time.
So, definitely not gonna be A.
B, aldehyde.
So, I see my aldehyde right here.
Right, 'cause an aldehyde group
is basically this right here.
However, when I look over here, I do not see it.
I just see, the closest thing I see is this right here,
but that's a carboxylic acid group,
and so carboxylic acid is one of the other answer choices.
And, you know, it is present here in the reactants
but it's also present in the product.
So, we can actually use that information
to eliminate answer choice C.
And, so, still really thinking
it's probably gonna be aldehyde.
Let's look at answer choice D.
They say amines here, so I see that right here.
And, of course, I see amines in the products, as well.
So, it's not D.
It looks like B, again, is our best answer.
In case you're not familiar with amines.
Amines, they come in a primary, secondary, tertiary.
A primary amine would be something like this,
a secondary would be this,
and then a tertiary would be this.
So, if it has no hydrogens, it's gonna be tertiary,
one hydrogen, secondary,
and then two hydrogens is gonna be primary.
And, then, in all of these,
the nitrogen has one lone pair here.
So, something you should know for the MCAT,
just like amino acids
are a very high-yield biochemistry topic,
your functional groups
are one of the highest-yield topics, as well.
And, knowing your functional groups
can actually save you a lot of time.
If you just know how these different functional groups react
and how they behave, then there's a lot of other material
you don't really have to memorize,
like certain reactions,
because you'll know the fundamentals of functional groups
and that can guide your, you know, problem solving
when you're working with very complicated reactions.
So, this question here,
if you weren't able to get the right answer,
I would say really review your functional groups.
Alrighty, so let's go ahead and move on to question three.
All right, so they ask
which of the following urease mutations at position 87
is most likely to maintain pi-stacking interactions
with compound 8.
All right, so they mention
specifically pi-stacking interactions.
Pi-stacking interactions are formed between aromatic rings.
So, the answer choice that we choose,
whether it's alanine, proline, threonine, or phenylalanine,
needs to be one that has aromatic rings.
So, alanine is a pretty simple amino acid.
It's just basically CH3 for its side chain.
Proline is gonna look a little something like this.
It's a little bit more complicated,
but you have CH2 here, you'll have another CH2,
and then another CH2,
and then it ties in here to form this little structure,
and then this is kinda the rest of the amino acid,
like backbone here, and you'll have your O minus,
and then you'll have a hydrogen here.
So, this is not an aromatic ring.
You know, this is what an aromatic ring looks like.
You know, we can see them here in our Figure 1 here,
like this is an aromatic ring.
This would not be.
So it's not A, it's not B.
Threonine, except I'm running out of space,
maybe I'll write it.
Let's put that this was answered choice A, this was B.
So, threonine, it looks something like this.
You got the rest of the backbone kind of up here
and then you have a carbon, you got this hydroxyl,
a methyl group here,
and this carbon has also gotten to an oxygen.
So, this is threonine here.
Again, we don't see an aromatic ring,
so we can eliminate C.
That leaves us with D
and it is correct because phenylalanine,
and since I've run out of space here, I'm gonna,
I'll just write it up here.
Phenylalanine looks something like this,
CH2 and then this little aromatic ring.
And, so, with that, having the aromatic ring,
it means it's gonna be the one
that's most likely to maintain the pi-stacking interactions.
So, D is our best answer.
Alrighty, let's go over the last question.
Researchers want to decrease the rate of product formation
from the second step of the reaction, Figure 1.
Which the following would, again, not decrease,
let's go ahead and underline that or highlight it,
would not decrease the rate of compound two formation
according to Le Chatelier's principle.
All alright, so Le Chatelier's principle
is just basically this idea where
when a stress is placed on a reaction,
it's gonna shift in a way to alleviate that stress.
So, these are the options they give us.
A, decrease the concentration
of the bromine-containing initial reactant.
All right, so if we were to decrease,
so this is our bromine-containing reactant,
if we decrease the concentration of this,
what you would have is a shift towards this way,
towards the reactants here, right?
So, because this is going down,
we have less material or less, you know,
if you kinda wanna look at this as, you know,
kinda balancing the reaction
in the sense that if we have less of this now,
that means our reaction is gonna shift this way
to restore the balance here.
So, we would have more of our initial reaction forming here
and we would also have compound two, right here,
its formation would decrease
'cause there's less of this starting material.
So, summarize, with this going down,
the reaction would shift this way.
So, A is not gonna be the right answer
because question asks us which one does not decrease
the rate of compound two formation,
and A does decrease the rate of compound two formation.
All right, B, decrease the concentration of
aldehyde- and carboxylic acid-containing initial reactant.
All right, so this right here is our
aldehyde- and carboxylic acid-containing initial reactant,
and the same logic as our justification for why A is wrong
applies here for why B is gonna be wrong.
If this is decreased, then this thing, this reaction
is gonna shift that way.
And, again, we're decreasing the formation of compound two.
Question asks us
which doesn't decrease that rate of formation.
Since B does, we can eliminate it.
C, decrease the concentration of compound two
All right, so take that same logic
we've been applying with A and B and see what happens here.
If, when we lower, like, let's say for answer choice A,
when we lowered this reactant here,
our reaction shifted to restore the balance
and decreased the rate of formation for the products here.
So, if we're doing kinda the same thing now
but with compound two, right,
if we're decreasing compound two's concentration,
then we would have this reaction actually shift
to restore that balance
and increase that concentration of compound two.
So, this one would increase
the concentration of compound two
or the rate of compound two formation.
So, it would not, this would be, you know,
the one that does not decrease
the rate of compound two formation.
So, I really like this as an answer choice.
Let's go ahead and look at answer choice D
before we lock in C.
Decrease the concentration of compound one.
So, again, apply the same logic we've used for A, B and C
and look at compound one here.
If we decrease this concentration,
then we would have this part of the reaction
shifting to the right.
Right, in our first step, we would have this shift right
and, in our second step, the compound two, right,
this reaction here would actually shift left here.
So, just looking at this reaction here,
if we're decreasing the concentration of our reactant
that forms the product of compound two,
then we're gonna get more reactant formation
and less of our compound two or product formation.
So, D does decrease the rate of compound two formation,
making it incorrect.
And, so, C is our best to answer.
All right, so that's gonna wrap things up for us today.
So, the key things really being
if you weren't able to get question one,
definitely review your strong acids.
For question two, definitely review the functional groups.
Three, your amino acids are huge, right?
And, for four, review Le Chatelier's principle.
These are all high-yield topics for test day,
so make sure that if you did miss any of these questions
that you go back to your content review books
or check out one of our MCAT review content guides
on these topics that you might've struggled with.
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Have a great one and happy studying.
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